Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}4x+2y &= -8 \\ -6x+y &= -3\end{align*}$
Answer: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-6x = -y-3$ Divide both sides by $-6$ to isolate $x$ $x = {\dfrac{1}{6}y + \dfrac{1}{2}}$ Substitute this expression for $x$ in the first equation. $4({\dfrac{1}{6}y + \dfrac{1}{2}}) + 2y = -8$ $\dfrac{2}{3}y + 2 + 2y = -8$ Simplify by combining terms, then solve for $y$ $\dfrac{8}{3}y + 2 = -8$ $\dfrac{8}{3}y = -10$ $y = -\dfrac{15}{4}$ Substitute $-\dfrac{15}{4}$ for $y$ in the top equation. $4x+2( -\dfrac{15}{4}) = -8$ $4x-\dfrac{15}{2} = -8$ $4x = -\dfrac{1}{2}$ $x = -\dfrac{1}{8}$ The solution is $\enspace x = -\dfrac{1}{8}, \enspace y = -\dfrac{15}{4}$.